Is it possible for 2 + 2 = 5?

Answer 1

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You can prove it by de-unifying the number scale, that is, dividing by zero, since any number divided by zero is equal to any other number.

But in REAL math, it is impossible.

Here's the divide by zero thing:

0 * (2+2) = 0 * (5) (true)

(0 * (2+2)) / 0 = (0 * (5)) / 0 (divide both sides by zero, and the zeros cancel out)

(2+2) = (5) (so it looks like they equal each other)

Answer 2

It's possible for 2 + 2 = 5 if you change standard conventions.

If you assume a false proposition, you can deduce anything from it and make it true, for one, so it's logically admissible to say "all cows fly, therefore 2 + 2 = 5".

Also, you can change the meaning of the symbols 2, +, = and 5 so they represent things that will make the statement true. For example, let = be still =, let 2 still be 2, let + mean / and let 5 mean 1. I shouldn't need to say there are an infinity of ways to do this.

Finally, you can suppose it's admissible to divide by cero in some system, which would mean you're allowed to say 1 = 0, so it's easy to deduce that 2 + 2 = 5.

Had fun yet? Unless you can do something useful with 2 + 2 = 5, it's kinda silly to try and force it to be that way. Oh, and by the way, this is a maths question, I don't think physics can answer it.

Source(s): I'm studying to become a mathematician.

Answer 3

It could be, depending on how you define 2, + ,= , and 5.

What if that 5 became our current 4? So that counting up yields 1,2,3,5,4. There's really nothing stopping you from using this convention besides confusing other people. You can get even crazier and say that + is actually what is known as the subtraction operator, 5 is actually 0 and so 2 + 2 = 5 becomes equivalent to 2 - 2 = 0. It all depends on how you define it.

After all, they're just symbols to an underlying meaning.

Answer 4

it somewhat is a sort of programming code, yet incomplete. it potential whilst (2+2 = real) then 5, a variable gets the cost 5 if 2+2 = real, in Perl is any term which has the effect >=a million real and the effect 0=fake, so 2+2 = real (in Perl) then a variable which you probably did no longer point out gets the cost 5

Answer 5

Yes.

Source(s): Magic.

Answer 6

Unless this is somekind of trick question,

No.

Source(s): MPhys

Answer 7

possibly, who knows?

anything could be possible.

anything.

Answer 8

Yes...

I am sure you can prove this with a very lengthy prove. I can't...

Answer 9

Not yet anyway!

Answer 10

not really unless u are blind